# Proving that the $\lfloor-x\rfloor= -\lceil x\rceil$

My homework assignment has asked me to prove that $$\lfloor-x\rfloor = -\lceil x\rceil$$.

Conceptually this makes perfect sense to me, I just am at a loss for how to start actually proving it. I figure that it has something to do with these definitions: $$\lfloor x \rfloor: a \le x \lt a+1$$ $$\lceil x \rceil : a \lt x \le a+1$$

But I'm unsure of how to apply these to devise a proof for this. Any help that anyone could give would be incredibly appreciated, I've spent way too much time staring at the question trying to figure it out.

## 8 Answers

$$\begin{eqnarray}-x-1< &\lfloor -x \rfloor &\leq -x \\ x\leq &\lceil x \rceil &< x+1 \end{eqnarray}$$

Adding these two we get $$(-x-1)+x< \lfloor -x \rfloor + \lceil x \rceil <-x+(x+1)$$

$$-1< \lfloor -x \rfloor + \lceil x \rceil <1$$ so $$\lfloor -x \rfloor + \lceil x \rceil =0$$

• Very nice proof! – mjw Mar 26 at 21:47

Hint

$$k\le -x

If $$x$$ is an integer than an obvious equality holds. Otherwise we have $$k-1\lt x\lt k$$ for some $$k\in\mathbb{Z}$$. So we have $$\lfloor -x\rfloor=-k$$ $$-\lceil x \rceil = -k$$ which are equal. The first equality is true because $$-k\lt-x\lt1-k$$ so $$\lfloor -x\rfloor$$ takes the lower value. The second is true as $$\lceil x \rceil$$ takes the higher value and it becomes negated.

The left-hand side says ($$p_1=\lfloor{-x}\rfloor) \quad$$ $$-x=p_1+\varepsilon$$ for some $$0 \le \varepsilon \le 1$$.

The right-hand side says ($$p_2=-\lceil{x}\rceil)\quad$$ $$p_2=-(x+\delta)$$ for some $$0 \le \delta \le 1$$.

Since both $$\varepsilon$$ and $$\delta$$ are the distances from $$-x$$ to nearest integer less than $$x$$ they are equal and $$p_1 = p_2$$.

With most elementary floor and ceiling proofs, the way to start is: Let $$x = k +r$$ where $$k$$ is an integer and $$0\leq r <1.$$ That usually fixes everything.

You have, if $$r\neq 0$$:

$$\lfloor -x \rfloor = \lfloor -k-r \rfloor = -k +\lfloor -r \rfloor = -k -1.$$

and:

$$-\lceil x \rceil = -\lceil k+r \rceil = -(k+\lceil r \rceil) = -k -1.$$

And if $$r=0$$, a similar thing happens.

Just do it:

We know that $$\lfloor -x \rfloor \le -x < \lfloor -x \rfloor + 1$$ and that $$\lfloor -x \rfloor$$ is an integer.

So $$-\lfloor -x \rfloor \ge x > -\lfloor -x \rfloor-1$$ and so $$-\lfloor -x \rfloor- 1< x \le -\lfloor -x \rfloor$$ and $$-\lfloor -x \rfloor$$ is an intger.

So by definition that means $$\lceil x \rceil = -\lfloor -x \rfloor$$.

And $$\lfloor -x \rfloor = -\lceil x \rceil$$

Let $$x \in \mathbb{R}$$ be given. By definition, $$\lfloor -x \rfloor$$ is the unique integer $$a$$ such that $$a \leq -x < a+1.$$ Now, this implies that $$-(a+1) < x \leq -a$$ where $$-a \in \mathbb{Z}$$ also. By definition of $$\lceil x \rceil$$, we must then have $$\lceil x \rceil = -a$$. In short, we have proven that \begin{align*} \lfloor -x \rfloor = a = -\lceil x \rceil. \end{align*}

The map $$\varphi\colon x\mapsto -x$$ is an order isomorphism of $$(\mathbb{R},\le)$$ onto $$(\mathbb{R},\ge)$$.

Since $$\lfloor x\rfloor=\sup_{\le}\{z\in\mathbb{Z}:z\le x\}$$ and $$\varphi$$ maps $$\mathbb{Z}$$ onto $$\mathbb{Z}$$, you're done, taking into account that $$\lceil x\rceil=\sup_{\ge}\{z\in\mathbb{Z}:z\ge x\}$$ and, obviously, $$\sup_{\ge}$$ is the same as $$\inf_{\le}$$.

• Is $\lceil x \rceil$ a supremum or an infimum? – mjw Mar 27 at 15:43
• @mjw It's a supremum with respect to $\ge$ and an infimum with respect to $\le$. – egreg Mar 27 at 15:46
• I would have expected $\lceil x \rceil$ and $\lfloor x \rfloor$ to have dual descriptions. – mjw Mar 27 at 15:48
• @mjw They have: one is a supremum with respect to $\ge$, the other one is a supremum with respect to $\le$. Do you see the duality? – egreg Mar 27 at 15:49
• Yes, I would have thought, though, that $\lceil x\rceil=\inf_{\ge}\{z\in\mathbb{Z}:z\ge x\}$. – mjw Mar 27 at 15:51